Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int L[100010];
int P[100010][20];
vector<int> pp[100010][20];
int T[100010];
vector<int> tt[100010];
int arr[100010];
vector<int> graph[100010];
vector<int> fnc(vector<int> &a, vector<int> &b) {
  vector<int> ans;
  int aa = a.size(), bb = b.size(), i, j, shajib = 0;
  for (i = 0, j = 0; i + j < aa + bb && shajib < 10;) {
    if (i == aa) {
      ans.push_back(b[j]);
      j++;
      shajib++;
    } else if (j == bb) {
      ans.push_back(a[i]);
      i++;
      shajib++;
    } else {
      if (a[i] == b[j]) {
        i++;
      } else if (a[i] > b[j]) {
        ans.push_back(b[j]);
        j++;
        shajib++;
      } else {
        ans.push_back(a[i]);
        i++;
        shajib++;
      }
    }
  }
  return ans;
}
void dfs(int src, int par) {
  if (src == -1)
    L[par] = 1;
  else
    L[par] = L[src] + 1;
  T[par] = src;
  int len = graph[par].size();
  for (int i = 0; i < (int)graph[par].size(); i++) {
    int v = graph[par][i];
    if (v != src) {
      dfs(par, v);
    }
  }
}
vector<int> query(int n, int p, int q) {
  int log = 1, i, j, temp;
  if (L[p] < L[q]) {
    temp = q;
    q = p;
    p = temp;
  }
  log = log2(n);
  vector<int> os;
  os = fnc(os, tt[p]);
  for (j = log; j >= 0; j--) {
    if (L[p] - (1 << j) >= L[q]) {
      os = fnc(os, pp[p][j]);
      p = P[p][j];
    }
  }
  if (p == q) return os;
  os = fnc(os, tt[q]);
  for (j = log; j >= 0; j--) {
    if (P[p][j] != -1 && P[p][j] != P[q][j]) {
      os = fnc(os, pp[p][j]);
      os = fnc(os, pp[q][j]);
      p = P[p][j], q = P[q][j];
    }
  }
  os = fnc(os, pp[p][0]);
  return os;
}
void lca(int n) {
  memset(P, -1, sizeof(P));
  int i, j;
  for (i = 1; i <= n; i++) {
    P[i][0] = T[i];
    if (T[i] != -1) pp[i][0] = fnc(pp[i][0], tt[T[i]]);
  }
  for (j = 1; (1 << j) < n; j++) {
    for (i = 1; i <= n; i++) {
      if (P[i][j - 1] != -1) {
        P[i][j] = P[P[i][j - 1]][j - 1];
        pp[i][j] = fnc(pp[i][j], pp[P[i][j - 1]][j - 1]);
        pp[i][j] = fnc(pp[i][j], pp[i][j - 1]);
      }
    }
  }
}
int main() {
  int i, j, n, m, t, a, b, k, c;
  scanf("%d%d%d", &n, &m, &k);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  for (i = 1; i <= m; i++) {
    scanf("%d", &a);
    tt[a].push_back(i);
  }
  for (i = 1; i <= n; i++) {
    sort(tt[i].begin(), tt[i].end());
    a = tt[i].size();
    while (a > 10) {
      tt[i].pop_back();
      a--;
    }
  }
  dfs(-1, 1);
  lca(n);
  vector<int> os;
  for (i = 1; i <= k; i++) {
    scanf("%d%d%d", &a, &b, &c);
    os = query(n, a, b);
    c = min(c, (int)os.size());
    a = 0;
    printf("%d ", c);
    for (auto it = os.begin(); a < c && it != os.end(); a++, it++) {
      printf("%d ", (*it));
    }
    printf("\n");
  }
}